Asked by Suman Koirala
on 23 Apr 2013

Quiz 9: Calculation of Production Mix

Hours

Reactor A

Reactor B

Product 1

5

3

Product 2

3

3

Product 3

3

4

The table shows how many hours it takes reactors A and B to produce a ton of each

product. From the table we see that for every ton of product 1 produced we need to

run Reactor A for 5 hours and also run Reactor B for 3 hours. The two reactors are

available for 40 hours and 30 hours per week respectively. We can determine the

amount (in tons) of product 1, p 1 , product 2, p 2 , and product 3, p 3 that can be

produced in a week using the following equations:

5 p 1 + 3 p 2 + 3 p 3 = 40

3 p 1 + 3 p 2 + 4 p 3 = 30

Let F 1 , F 2 , F 3 represent the profit for a ton of each product . So the total weekly

profit is given by:

Profit = F 1 p 1 + F 2 p 2 + F 3 p 3

Create a script M-file that uses the reduced row echelon function rref to solve for p 1

and p 2 as functions of p 3 , and generates two subplots. The first subplot should show

the total weekly profit as a function of p 3 for the following two cases:

Case 1: F 1 = $400, F 2 = $600, F 3 = $100

Case 2: F 1 = $600, F 2 = $200, F 3 = $200

The second subplot shows p 1 and p 2 as a function of p 3 .

Based on the results of the first subplot, your production manager says that she is

going to adopt an all or nothing strategy on the production volume of product 3. She

says that the plant will either produce the maximum amount of product 3 or none at

all. What is her justification for this?

This is what I have so far. My second subplot looks wierd.

A=[5 3; 3 3]; b=[40- 3*p3; 30-4*p3]; p=A\b;

f1=400; f2=600; f3=100; p3=[0:0.1:40]; P=f1*p(1)+f2*p(2)+f3*p3; subplot(1,2,1) plot(p3, P) hold on f1=600; f2=200; f3=200; P=f1*p(1)+f2*p(2)+f3*p3; plot(p3, P)

subplot(1,2,2) plot(p(1),p3) hold on plot(p(2),p3)

Any suggestion is appreciated. Thanks!

*No products are associated with this question.*

Related Content

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today
## 0 Comments

Log in to comment.